calorimetry: final temperature of the solution formed after NaCl is dissolved in water

Question

consider the following reaction:

NaCl (s) -> Na+ (aq) + Cl- (aq) ΔH=3.88 kJ/mol

In a coffee-cup calorimeter you dissolve 23.3 g of this salt in 117 g of water at 27.7 C. what will be the final temperature of the solution formed?

assume the specific heat of the solution is 4.184 J/gC, and that there is no heat loss to the surroundings

Answer 1

heat = mass x specific heat x change in temperature

q = mC∆T

m = 117 g water

C = 4.184 J/g/deg

∆T = ?

q = heat, but to find this we need to first find moles NaCl used in this reaction:

23.3 g NaCl x 1 mole/58.44 g = 0.399 moles

q = 0.399 moles x 3.88 kJ/mol = 1.55 kJ = 1550 J (since it is positive, this indicates an endothermic reaction, so temperature of solution will decrease)

1550 J = (117g)(4.184 J/g/deg)(∆T)

∆T = 3.17 degrees

Final temperature = 27.7º - 3.17º = 24.5ºC